Solving a linear system of equations in 3 unknowns











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I am having trouble solving this linear system of equations to get $theta_1$ and $theta_2$ as functions of $x$ and $y$ where $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are constants:



$theta_1cdot x_1 + theta_2cdot x_2 + theta_3 cdot x_3 = x$



$theta_1cdot y_1 + theta_2cdot y_2 + theta_3 cdot y_3 = y$



$theta_1 + theta_2 + theta_3 = 1$



By using the condition $theta_3 = 1-theta_1 - theta_2$ i have got to the following two-by-two system of equations:



$theta_1(x_1-x_3) + theta_2(x_2-x_3) = x - x_3$



$theta_1(y_1-y_3) + theta_2(y_2-y_3) = y - y_3$



Can anyone help me as to how to proceed from here? Also, I have just been using brute-force so far and would like to know if there is an easier or more sophisticated way of solving this system as the above two equations look very similar.



Any help would be greatly appreciated, thanks guys!






linear-algebra matrices numerical-methods systems-of-equations matrix-equations







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  • You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
    – LutzL
    Nov 21 at 17:30















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I am having trouble solving this linear system of equations to get $theta_1$ and $theta_2$ as functions of $x$ and $y$ where $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are constants:



$theta_1cdot x_1 + theta_2cdot x_2 + theta_3 cdot x_3 = x$



$theta_1cdot y_1 + theta_2cdot y_2 + theta_3 cdot y_3 = y$



$theta_1 + theta_2 + theta_3 = 1$



By using the condition $theta_3 = 1-theta_1 - theta_2$ i have got to the following two-by-two system of equations:



$theta_1(x_1-x_3) + theta_2(x_2-x_3) = x - x_3$



$theta_1(y_1-y_3) + theta_2(y_2-y_3) = y - y_3$



Can anyone help me as to how to proceed from here? Also, I have just been using brute-force so far and would like to know if there is an easier or more sophisticated way of solving this system as the above two equations look very similar.



Any help would be greatly appreciated, thanks guys!






linear-algebra matrices numerical-methods systems-of-equations matrix-equations







share|cite|improve this question






















  • You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
    – LutzL
    Nov 21 at 17:30













up vote
0
down vote

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up vote
0
down vote

favorite











I am having trouble solving this linear system of equations to get $theta_1$ and $theta_2$ as functions of $x$ and $y$ where $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are constants:



$theta_1cdot x_1 + theta_2cdot x_2 + theta_3 cdot x_3 = x$



$theta_1cdot y_1 + theta_2cdot y_2 + theta_3 cdot y_3 = y$



$theta_1 + theta_2 + theta_3 = 1$



By using the condition $theta_3 = 1-theta_1 - theta_2$ i have got to the following two-by-two system of equations:



$theta_1(x_1-x_3) + theta_2(x_2-x_3) = x - x_3$



$theta_1(y_1-y_3) + theta_2(y_2-y_3) = y - y_3$



Can anyone help me as to how to proceed from here? Also, I have just been using brute-force so far and would like to know if there is an easier or more sophisticated way of solving this system as the above two equations look very similar.



Any help would be greatly appreciated, thanks guys!






linear-algebra matrices numerical-methods systems-of-equations matrix-equations







share|cite|improve this question













I am having trouble solving this linear system of equations to get $theta_1$ and $theta_2$ as functions of $x$ and $y$ where $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are constants:



$theta_1cdot x_1 + theta_2cdot x_2 + theta_3 cdot x_3 = x$



$theta_1cdot y_1 + theta_2cdot y_2 + theta_3 cdot y_3 = y$



$theta_1 + theta_2 + theta_3 = 1$



By using the condition $theta_3 = 1-theta_1 - theta_2$ i have got to the following two-by-two system of equations:



$theta_1(x_1-x_3) + theta_2(x_2-x_3) = x - x_3$



$theta_1(y_1-y_3) + theta_2(y_2-y_3) = y - y_3$



Can anyone help me as to how to proceed from here? Also, I have just been using brute-force so far and would like to know if there is an easier or more sophisticated way of solving this system as the above two equations look very similar.



Any help would be greatly appreciated, thanks guys!






linear-algebra matrices numerical-methods systems-of-equations matrix-equations




linear-algebra matrices numerical-methods systems-of-equations matrix-equations






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asked Nov 21 at 17:14









Hallao

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  • You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
    – LutzL
    Nov 21 at 17:30


















  • You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
    – LutzL
    Nov 21 at 17:30
















You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
– LutzL
Nov 21 at 17:30




You are trying to compute the barycentric coordinates of $(x,y)$ relative to the triangle spanned by the $(x_k,y_k)$. What exactly is the problem of inverting the $3times 3$ matrix? Or solve the reduced $2times 2$ system? No one is saying that the solution has to look nice, or at least nicer than the Cramer solution formulas.
– LutzL
Nov 21 at 17:30










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You can rewrite your system of equations in the following form:
$$begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x \ y \ 1 end{bmatrix}$$



If the matrix containing the $x_i$ and $y_i$ is invertible, you can find a unique solution for your $theta_i$ as:



$$begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}^{-1}begin{bmatrix} x \ y \ 1 end{bmatrix}$$



which you can calculate either directly or by using numerics. If your matrix is not invertible, then you do not have a unique solution and may have either no solution at all or an infinite number of solutions depending on the image of the matrix containing the $x_i$'s and $y_i$'s.






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    You can rewrite your system of equations in the following form:
    $$begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x \ y \ 1 end{bmatrix}$$



    If the matrix containing the $x_i$ and $y_i$ is invertible, you can find a unique solution for your $theta_i$ as:



    $$begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}^{-1}begin{bmatrix} x \ y \ 1 end{bmatrix}$$



    which you can calculate either directly or by using numerics. If your matrix is not invertible, then you do not have a unique solution and may have either no solution at all or an infinite number of solutions depending on the image of the matrix containing the $x_i$'s and $y_i$'s.






    share|cite|improve this answer

























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      You can rewrite your system of equations in the following form:
      $$begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x \ y \ 1 end{bmatrix}$$



      If the matrix containing the $x_i$ and $y_i$ is invertible, you can find a unique solution for your $theta_i$ as:



      $$begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}^{-1}begin{bmatrix} x \ y \ 1 end{bmatrix}$$



      which you can calculate either directly or by using numerics. If your matrix is not invertible, then you do not have a unique solution and may have either no solution at all or an infinite number of solutions depending on the image of the matrix containing the $x_i$'s and $y_i$'s.






      share|cite|improve this answer























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        You can rewrite your system of equations in the following form:
        $$begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x \ y \ 1 end{bmatrix}$$



        If the matrix containing the $x_i$ and $y_i$ is invertible, you can find a unique solution for your $theta_i$ as:



        $$begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}^{-1}begin{bmatrix} x \ y \ 1 end{bmatrix}$$



        which you can calculate either directly or by using numerics. If your matrix is not invertible, then you do not have a unique solution and may have either no solution at all or an infinite number of solutions depending on the image of the matrix containing the $x_i$'s and $y_i$'s.






        share|cite|improve this answer












        You can rewrite your system of equations in the following form:
        $$begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x \ y \ 1 end{bmatrix}$$



        If the matrix containing the $x_i$ and $y_i$ is invertible, you can find a unique solution for your $theta_i$ as:



        $$begin{bmatrix} theta_1 \ theta_2 \ theta_3 end{bmatrix} = begin{bmatrix} x_1 & x_2 & x_3\ y_1 & y_2 & y_3\ 1 & 1 & 1 end{bmatrix}^{-1}begin{bmatrix} x \ y \ 1 end{bmatrix}$$



        which you can calculate either directly or by using numerics. If your matrix is not invertible, then you do not have a unique solution and may have either no solution at all or an infinite number of solutions depending on the image of the matrix containing the $x_i$'s and $y_i$'s.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 20:26









        aghostinthefigures

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